Suppose $r$ and $s$ are the values of $x$ that satisfy the equation \[x^2 - 2mx + (m^2+2m+3) = 0\]for some real number $m$. Find the minimum real value of $r^2+s^2$.
First, note that by Vieta's formulas, r^2 + s^2 will be a polynomial in m. Since m is real, we know r^2 + s^2 will be real as well. So, the problem is simply asking us to find the minimum value of r^2 + s^2.
By Vieta's formulas, we have
\(r+s=2m \quad\text{and}\quad rs=m^2+2m+3\)
Therefore,
\(\begin{align*} r^2+s^2 &= (r+s)^2 - 2(rs) \\ &= (2m)^2 - 2(m^2+2m+3) \\ &= 2m^2-4m-6 \\ &= 2(m^2-2m-3). \end{align*}\)
Completing the square gives
\(r^2+s^2 = 2\left((m-1)^2-4\right).\)
This is minimized when m=1, and the minimum value is \(\boxed{-8}\).
First, note that by Vieta's formulas, r^2 + s^2 will be a polynomial in m. Since m is real, we know r^2 + s^2 will be real as well. So, the problem is simply asking us to find the minimum value of r^2 + s^2.
By Vieta's formulas, we have
\(r+s=2m \quad\text{and}\quad rs=m^2+2m+3\)
Therefore,
\(\begin{align*} r^2+s^2 &= (r+s)^2 - 2(rs) \\ &= (2m)^2 - 2(m^2+2m+3) \\ &= 2m^2-4m-6 \\ &= 2(m^2-2m-3). \end{align*}\)
Completing the square gives
\(r^2+s^2 = 2\left((m-1)^2-4\right).\)
This is minimized when m=1, and the minimum value is \(\boxed{-8}\).
